Which equation has the same solution as x^2-6x-12=0?

1) (x+3)^2=21 2) (x-3)^2=2

3) (x+3)^2=3 4) (x-3)^2=3

                  x² - 6x - 12 = 0
                            + 12 + 12
                         x² - 6x = 12
                   x² - 6x + 9 = 12 + 9
            x² - 3x - 3x + 9 = 21
x(x) - x(3) - 3(x) + 3(3) = 21
         x(x - 3) - 3(x - 3) = 21
                (x - 3)(x - 3) = 21
                        (x - 3)² = 21

The answer is 2.

I see that you complete the square

basically
for
ax^2+bx+c=0
isolate the x terms
(ax^2+bx)+c=0
factor out a if a is not equal to 1
take 1/2 of b and square it and add posotive and negative inside

so

x^2-6x-12=0
move c aside
(x^2-6x)-12=0
first term iss 1 do don't undistribute
take 1/2 of -6 and square it (-6/2=-3, (-3)^2=9)
add negative and positive
(x^2-6x+9-9)-12=0
factor perfect square
((x-3)^2-9)-12=0
(x-3)^2-9-12=0
(x-3)^2-21=0
add 21 both sides
(x-3)^2=21

yah, answer is 2nd one